The sum $10 e^{2 \pi i/11} + 10 e^{15 \pi i/22}$ is expressed as $re^{i \theta}.$  Enter the ordered pair $(r, \theta).$
Solution: The average of $\frac{2 \pi}{11}$ and $\frac{15 \pi}{22}$ is $\frac{19 \pi}{44}.$  We can then write
\begin{align*}
10 e^{2 \pi i/11} + 10 e^{15 \pi i/22} &= 10 e^{19 \pi i/44} (e^{-\pi i/4} + e^{\pi i/4}) \\
&= 10 e^{19 \pi i/44} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} + \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right) \\
&= 10 \sqrt{2} e^{19 \pi i/44}.
\end{align*}Thus, $(r, \theta) = \boxed{\left( 10 \sqrt{2}, \frac{19 \pi}{44} \right)}.$